Question 72
Let $A_{1}$, $B_{1}$ and $C_{1}$ be respectively the midpoints of the sides $BC$, $CA$ and $AB$ of $\triangle ABC$ with centroid $M$. The line through $A_{1}$ and parallel to $BB_{1}$ meets the line $B_{1}C_{1}$ at a point $D$. Prove that if the points $A$, $B_{1}$, $M$ and $C_{1}$ are concyclic, then $\measuredangle ADA_{1} = \measuredangle CAB$.